A few administrative notes before we review the day’s material: I will not be holding office hours this Wednesday. And there are no classes next Monday, when my usual set of office hours are. But I’ve decided to do a sort of experiment: I don’t plan on reviewing for the exam specifically next week, but a large portion of the class has said that they would come to office hours on Monday if I were to have them. So I’m going to hold them to that – I’ll be in Kassar House 105 (the MRC room) from 7-8:30 (or so, later perhaps if there are a lot of questions), and this will dually function as my office hours and a sort of review session.

But this comes with a few strings attached: firstly, I’ll be willing to answer any question, but I’m not going to prepare a review; secondly, if there is poor turnout, then this won’t happen again. Alrighty!

The rest is after the fold –

The topic of the day was differentiation! The three questions of the day were –

- Differentiate the following functions:
- A particle moves along a line with its position described by the function . If we know that it’s acceleration is always m/s/s, that its velocity at is m/s, and its position at is m. What are ?
- Given that , what are the following:

#### Question 1

This is all about the chain rule. Please note that this is a big deal, so if you have any trouble at all with the chain rule, seek extra help. The derivative of is . To compute the derivative of , we might think of , so that we have . The derivative of will be , which gives us . Let’s look at the other way of understanding the chain rule to compute the derivative of . The “outer function” is . It’s derivative is just itself. The first “inner function” is . We have just computed its derivative above (it’s ). So we multiply them together to get .

Similarly, the derivative of is $\cos x$. The derivative of $\sin \sin x$ requires the chain rule. On the one hand, the outer function is , and the derivative of is $\cos$. So we know we will have a in the answer. The inner function is also , so we need to multiply by its derivative. The final answer will be . To compute the derivative of , we again use the chain rule. I will again use helper functions, to illustrate their use. We might call , so that we are computing the derivative of . Then we get . We happen to have computed just a moment ago, so the final answer is .

#### Question 2

The key idea of this question is to remember that the function gives position at time . So its derivative gives a result in terms of position per time, the velocity. And the derivative of velocity will give a result in terms of position per time per time, or acceleration. So the velocity of our particle is , and the acceleration is . Since we know that the acceleration is always , we know that so that . The velocity at is , so we know that , so that . Finally, our position at time is , so that , so that . I used different numbers between the two classes, so don’t pay too much attention if the exact details are different between one class and the other.

#### Question 3

This is more about the chain-rule! This is sort of an explicit example of helper functions. We first want to compute the derivative of . By the chain rule, this will be . What is ?. It’s . So the derivative of is . This is a single case of the slightly more general . Here, the power rule tells us that the derivative will be , which is .

The idea behind the third question is to see if we can work out the same sort of idea, but without starting with a helper function. (It’s perfectly fine to always use helper functions to use the chain rule – that’s not a problem at all). The derivative of will be . If we want to see the use of helper functions, call , so that we are computing the derivative of . The derivative will be , which is exactly what we have above.

I look forward to seeing some of you on Monday, and happy studying!

I’m writing on behalf of a emailed question that I feel might be appropriate for more people to see:

This comes in two parts.

Let’s find the points of continuity of . We know that , so we are wondering about when

To do this, we first need to remember when . The zeroes of are or for any integer (this includes , which you mentioned in your email). But we now have instead of $latex $, so we might ask ourselves when for any integer . This happens when .

Let’s check ourselves – . So this seems about right.

You ask: why do we only care about when is continuous when we consider ? To answer this, we need to remember something from the book:

If and are both continuous at a point , then their product is also continuous at the point .

For our problem, we might realize that is continuous everywhere, but is only continuous whenever for any integer . This means that their product is continuous at least when both and are continuous. So our function is continuous at least when .

But what about when ? Let’s look at one in particular: say . is not defined (the left limit is , the right limit is , but has a vertical asymptote there). In fact, $\tan x$ isn’t defined at any of these points , so these points are not in the domain of our function. So the function can’t be continuous there.

So the discontinuities of are exactly the places where is discontinuous.

During my office hours, I mentioned that this not because of anything particularly special about , but simply because is continuous. In a sense, multiplying by a continuous function never reduces the points of continuity.